1. 10.5 Moments of Inertia for
Composite Areas
A composite area consist of a series of
connected simpler parts or shapes such as
semicircles, rectangles and triangles
Provided the moment of inertia of each of
these parts is known or can be determined
about a common axis, moment of inertia of
the composite area = algebraic sum of the
moments of inertia of all its parts

2. 10.5 Moments of Inertia for
Composite Areas
Procedure for Analysis
Composite Parts
Using a sketch, divide the area into its composite
parts and indicate the perpendicular distance
from the centroid of each part to the reference
axis
Parallel Axis Theorem
Moment of inertia of each part is determined
about its centroidal axis, which is parallel to the
reference axis

3. 10.5 Moments of Inertia for
Composite Areas
Procedure for Analysis
Parallel Axis Theorem
If the centroidal axis does not coincide with the
reference axis, the parallel axis theorem is used
to determine the moment of inertia of the part
about the reference axis
Summation
Moment of inertia of the entire area about the
reference axis is determined by summing the
results of its composite parts

4. 10.5 Moments of Inertia for
Composite Areas
Procedure for Analysis
Summation
If the composite part has a hole, its moment
of inertia is found by subtracting the moment
of inertia of the hole from the moment of
inertia of the entire part including the hole

5. 10.5 Moments of Inertia for
Composite Areas
Example 10.5
Compute the moment of
inertia of the composite
area about the x axis.

6. 10.5 Moments of Inertia for
Composite Areas
Solution
Composite Parts
Composite area obtained
by subtracting the circle
form the rectangle
Centroid of each area is
located in the figure

7. 10.5 Moments of Inertia for
Composite Areas
Solution
Parallel Axis Theorem
Circle
2
I x = I x ' + Ad y
1
4
2 2
( )
= π (25) + π (25) (75) = 11.4 106 mm 4
4
Rectangle
2
I x = I x ' + Ad y
1
=
12
( )
(100)(150)3 + (100)(150)(75)2 = 112.5 106 mm 4

8. 10.5 Moments of Inertia for
Composite Areas
Solution
Summation
For moment of inertia for the composite
area,
I x = −11.4(106 ) + 112.5(10 6 )
( )
= 101 10 6 mm 4

9. 10.5 Moments of Inertia for
Composite Areas
Example 10.6
Determine the moments
of inertia of the beam’s
cross-sectional area
about the x and y
centroidal axes.

10. 10.5 Moments of Inertia for
Composite Areas
Solution
Composite Parts
Considered as 3
composite areas A, B, and
D
Centroid of each area is
located in the figure

11. 10.5 Moments of Inertia for
Composite Areas
Solution
Parallel Axis Theorem
Rectangle A
2
I x = I x ' + Ad y
1
=
12
( )
(100)(300)3 + (100)(300)(200)2 = 1.425 109 mm 4
2
I y = I y ' + Ad x
1
=
12
( )
(300)(100)3 + (100)(300)(250)2 = 1.90 109 mm 4

12. 10.5 Moments of Inertia for
Composite Areas
Solution
Parallel Axis Theorem
Rectangle B
2
I x = I x ' + Ad y
1
=
12
( )
(600)(100)3 = 0.05 109 mm 4
I y = I y ' + Ad x2
1
=
12
( )
(100)(600)3 = 1.80 109 mm 4

13. 10.5 Moments of Inertia for
Composite Areas
Solution
Parallel Axis Theorem
Rectangle D
2
I x = I x ' + Ad y
1
=
12
( )
(100)(300)3 + (100)(300)(200)2 = 1.425 109 mm 4
2
I y = I y ' + Ad x
1
=
12
( )
(300)(100)3 + (100)(300)(250)2 = 1.90 109 mm 4

14. 10.5 Moments of Inertia for
Composite Areas
Solution
Summation
For moment of inertia for the entire cross-
sectional area,
( ) ( ) ( )
I x = 1.425 109 + 0.05 109 + 1.425 109
= 2.90(10 )mm
9 4
I = 1.90(10 ) + 1.80(10 ) + 1.90(10 )
y
9 9 9
= 5.60(10 )mm
9 4